GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
#
Question
Type
1
Unit
Common Core State
Standard(s)
CE obj
DOK Level
Key
MC
G.SRT.10
6.15H
1
C
2
MC
G.SRT.10
6.15H
1
C
3
MC
G.SRT.11
6.16H
1
A
4
MC
G.SRT.11
6.16H
2
A
5
MTF
G.SRT.10
6.16H
1
B
6
MTF
G.SRT.10
6.16H
1
A
7
MTF
G.SRT.10
6.16H
1
B
8
MC
G.SRT.11
6.16H
1
A
9
MTF
G.SRT.11
6.16H
1
A
10
MTF
G.SRT.11
6.16H
1
B
11
MTF
G.SRT.11
6.16H
1
B
12
CR
G.SRT.11
6.16H
2
-
13
CR
G.SRT.11
6.16H
2
-
14
MC
G.SRT.9
6.18H
2
A
15
MC
G.SRT.9
6.18H
1
C
16
MC
3
G.GMD.3
7.11
1
C
17
MC
5
G.C.5
7.2
1
C
18
MC
5
G.C.5
7.2
1
C
19
MC
5
G.C.5
7.2
1
B
20
MC
3
G.GMD.1
7.4
2
C
21
CR
3
G.GMD.1
7.4
3
—
22
MC
5
G.C.5
7.5
2
C
23
MC
3
G.GMD.4
7.6
2
D
24
MC
3
G.GMD.3
7.8
1
D
25
MC
3
G.GMD.3
7.8
2
D
26
ER
3
G.GMD.3, G.MG.1
7.8
2
—
27
MC
3
G.GMD.3
7.11
1
C
28
MC
3
G.GMD.3
7.11
1
B
29
MC
3
G.GMD.3
7.11
2
B
30
MC
3
G.GMD.3
7.11
2
D
31
MC
3
G.MG.1
7.11
2
A
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
32
MC
3
G.GMD.4
7.13
2
A
33
MC
3
G.GMD.3
7.14
2
C
34
MC
3
G.GMD.1
7.16
1
A
35
MC
3
G.GMD.4
7.17
2
A
36
CR
3
G.GMD.3, G.GMD.4
7.17
2
—
37
MC
4
G.GPE.7
7.18
2
C
38
CR
4
G.GPE.5, G.GPE.7
7.18
2
—
39
CR
5
G.C.5, G.MG.1
7.18
2
—
40
CR
5
G.C.5, G.MG.1
7.18
2
—
41
CR
5
G.GPE.4
7.18
2
—
42
MC
3
G.MG.2
7.19
2
C
43
MC
3
G.MG.2
7.19
2
C
44
CR
3
G.GMD.3, G.MG.1
7.19
3
—
45
CR
3
G.GMD.3, G.MG.1, G.MG.2
7.19
2
—
46
CR
5
G.GPE.7
7.19
2
—
47
CR
5
G.C.1
8.1
3
—
48
MC
5
G.C.1
8.1
2
C
49
MTF
5
G.C.2
8.2
1
A
50
MTF
5
G.C.2
8.3
1
B
51
MC
5
G.C.3
8.5
1
A
52
MC
5
G.C.3
8.6
1
B
53
MC
5
G.C.3
8.6
1
A
54
MC
5
G.C.3
8.6
1
B
55
CR
5
G.C.3
8.6
2
—
56
MC
5
G.C.3
8.7
1
D
57
MC
5
G.C.3
8.7
1
D
58
CR
5
G.C.3
8.7
2
—
59
MC
5
G.C.2
8.9
1
B
60
MC
5
G.C.2
8.9
1
B
61
MC
5
G.C.2
8.9
2
B
62
MC
5
G.C.2
8.9
1
B
63
MC
5
G.C.2
8.9
1
B
64
MC
5
G.C.2
8.9
1
D
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
65
MC
5
G.C.2
8.9
1
C
66
MTF
5
G.C.2
8.9
1
A
67
MTF
5
G.C.2
8.10
1
B
68
MTF
5
G.C.2
8.10
1
A
69
MTF
5
G.C.2
8.10
1
A
70
MTF
5
G.C.2
8.10
1
B
71
MTF
5
G.C.2
8,10
1
B
72
CR
5
G.C.2
8.10
2
—
73
CR
5
G.GPE.5
8.10
2
—
74
MC
5
G.C.2
8.11
1
A
75
MC
5
G.C.2
8.11
2
C
76
MTF
5
G.C.2
8.11
1
A
77
MTF
5
G.C.2
8.11
1
B
78
MC
5
G.C.2
8.11
1
D
79
MC
5
G.C.2
8.11
1
C
80
MC
5
G.C.2
8.11
1
A
81
MC
5
G.C.2
8.11
1
B
82
MC
5
G.C.2
8.11
1
A
83
MC
5
G.C.2
8.11
2
B
84
MTF
5
G.C.2
8.11
1
B
85
MTF
5
G.C.2
8.11
2
A
86
MTF
5
G.C.2
8.11
2
B
87
MTF
5
G.C.2
8.11
2
A
88
MC
5
G.C.2
8.11
2
D
89
MC
5
G.C.2
8.11
1
D
90
MC
5
G.C.2
8.11
2
B
91
MC
5
G.C.2
8.11
1
C
92
MC
5
G.C.2
8.11
1
D
93
MC
5
G.C.2
8.11
1
B
94
CR
5
G.C.4
8.12H
2
—
95
CR
5
G.GPE.1
9.1
3
—
96
MC
5
G.GPE.1
9.2
2
D
97
MC
5
G.GPE.1
9.2
2
D
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
98
MC
5
G.GPE.1
9.2
1
D
99
MC
5
G.GPE.1
9.2
1
B
100
MC
5
G.GPE.1
9.3
2
A
101
MC
5
G.GPE.1
9.3
1
B
102
CR
4
G.GPE.1, G.GPE.4
9.4
3
—
103
MC
5
G.GPE.1
9.4
1
D
104
CR
5
G.C.5
9.5
2
—
105
MC
4
G.GPE.2
9.7
1
B
106
CR
4
G.GPE.2, G.GPE.4
9.8
3
—
107
CR
4
G.GPE.2
9.9
2
—
108
MC
6
S.CP.1
10.4
1
A
109
CR
6
S.CP.1, S.CP.2, S.CP.6, S.CP.7
10.6
1
—
110
MTF
6
S.CP.8
10.7H
2
A
111
MC
6
S.CP.8
10.7H
2
B
112
MC
6
S.CP.9
10.7H
2
A
113
MTF
6
S.MD.6
10.7H
2
B
114
MTF
6
S.MD.6
10.7H
2
A
115
MTF
6
S.MD.6
10.7H
2
B
116
MTF
6
S.MD.6
10.7H
2
A
117
MTF
6
S.CP.2
10.8
1
A
118
MTF
6
S.CP.1
10.9
1
A
119
MTF
6
S.CP.5
10.10
2
B
120
MTF
6
S.CP.3
10.11
2
B
121
MC
6
S.CP.3
10.11
2
D
122
MC
6
S.CP.5
10.12
1
A
123
MC
6
S.CP.7
10.13
1
C
124
CR
6
S.CP.3, S.CP.4, S.CP.5, S.CP.6
10.14
2
—
125
MC
6
S.CP.5
10.15
2
A
126
MC
6
S.CP.7
10.15
1
C
127
MC
6
S.CP.4
10.17
2
A
128
CR
6
S.CP.5, S.CP.6, S.CP.8, S.MD.7
10.18
2
—
129
MC
6
S.CP.9
10.19H
1
B
130
MC
6
S.CP.9
10.19H
1
D
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
12. (6.16) (HONORS) In triangle ABC, mB  25 , a = 6.2, and b = 4. Find all possible
measures of the remaining two angles and the third side.
This question assesses the student’s ability to use the law of sines. Using the law of sines,
sin A sin 25
sin114 sin 25
, so A  41 . C  114 .
so, c  8.6 .


6.2
4
c
4
There is a second possible solution, as this is an ambiguous case: A  139. C  16.
sin16 sin 25
, so c  2.6 .

c
4
13. (6.16) (HONORS) In triangle ABC, mB  25 , a = 6.2, and c = 4.
(a)
Find all possible measures of the remaining two angles and the third side.
(b)
Find all possible areas of the triangle
This question assesses the student’s ability to use the law of cosines, law of sines, and the
general formula for the area of the triangle.
(a) Using the law of cosines, b2  42  6.22  2  4 6.2 cos 25 , so b  3.1. Using the law
of sines,
sin C sin 25

, so C  33 , A  122 . .
4
3.1
(b) The area of the triangle is A 
1
 4  6.2  sin 25  5.2units 2
2
21. This question assesses the student’s understanding of the use of limiting principles to derive
formulas.
As the circle is cut into increasingly larger numbers of sectors and rearranged as shown, the
shape of the resulting figure becomes more like a parallelogram, where the base has a length
close to half the circle’s circumference and the height is its radius. The area of the parallelogram
is  r  r    r 2 .
r
r
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
26. This question assesses the student’s ability to apply volume formulas in modeling situations.
(a) The aquarium has a rectangular cross section 50 cm by 25 cm. The marbles have a diameter
of 1 cm, so an array of 50  25 marbles will cover the bottom of the tank. To fill the marbles
about 5 cm deep will take 50  25  5 = 6250 marbles. That is 6250/500 = 13 (12.5) bags of
marbles. (Note: if a student realizes that the marbles will probably not form a 3-D array
50  25  5, but have “layers” of different sizes, say 50  25 for the first, 49  24 for the second,
50  25 for the third, etc. this is acceptable. Precisely how the marbles are “packed” needs only
be reasonable.)
(b) The total volume of water and marbles in the tank is 50 cm  25 cm  27 cm = 33,750 cm3.
3
The volume of marbles in the tank is 6250    cm   3272.5 cm3 . So the volume of water is
3 2

3
30,477.5 cm or about 30.5 liters. (Treating the marbles as a solid 50 cm  25 cm  5 cm block
of glass is incorrect; there is space between the marbles. If a student makes the case that 6 layers
of marbles is closer to 5 cm tall than 5 layers, this is an acceptable solution.)
4
1
36. This question assesses the student’s ability to visualize threedimensional objects generated by rotations of two-dimensional objects
and the application of volume formulas for solids.
3
2
(a) The resulting shape is a frustum of a cone with a cylinder cut out of
the middle.
2
(b) The resulting volume would be the volume of a cone generated by the large triangle ( ABC )
minus the sum of the volume of the small cone (generated by AED ) and the inner cylinder.
1 2
1 2

 
Vlarge cone  3  r h  Vsmall cone  3  r h

 
1
1
2  
2

Vlarge cone    5  5  Vsmall cone    3 3
3
3

 
125
27

 
Vlarge cone  3
 Vsmall cone  3

 
Volume of solid =
125  27
 44

 18  
3
3
 3

Vinner cylinder   r 2 h
+
Vinner cylinder    3  2 
2
Vinner cylinder  18









GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
38. This question assesses the student’s to graph linear functions, construct parallel and/or
perpendicular lines to given lines on the coordinate plane, and compute the perimeter of a figure
using coordinate geometry.
(a) See graph.
(b) The line m has equation y = –2x + 20. See graph.
(c) See graph.
(d) AO = 10. OB = 5. BK  45  3 5 . KA  80  4 5 .
Perimeter = 15  7 5 .
39. This questions assesses the student’s ability apply the measurement of arc lengths.
s  2 r

360
 2  250000 miles 
 2181.661
0.50
360
miles
The approximate diameter of the moon is 2200 miles.
40. This questions assesses the student’s ability apply the measurement of arc lengths.
7.2
of the circumference of the earth. So,
360
360
 25000 miles .
the approximate circumference of the earth is 500 miles 
7.2
The arc length from A to B is given as 500 miles. It is
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
41. This questions assesses the student’s ability to apply the area of sectors of circles.
See diagram.
The dog’s area is three parts: ¾ of a circle of radius 9 feet, ¼ of a circle of radius 5 feet, and ¼ of
a circle of radius 3 feet.
3
1
1
2
2
2
A    9 ft     5 ft     3 ft 
4
4
4
A  69.25 ft 2
A  218 ft 2
44. This question assesses the student’s ability to apply volume formulas in modeling situations.
(a) The beaker is well approximated by a cylinder:
400 cm3    4 cm  h
2
h  8 cm
(b) The Erlenmeyer Flask well approximated by a cone:
1
2
400 cm3    4 cm  h
3
h  24 cm
(c) The Florence Flask is well approximated by a sphere:
4
400 cm3   r 3
3
r  4.6 cm
This is larger than the beaker’s radius, so it will NOT fit inside the beaker.
(d) As one moves upward from the bottom of the flask, the cross-sectional area
decreases. Therefore, greater height (of the cone) is needed to hold an equivalent
volume.
(e) The 200-ml mark must be roughly half way between the bottom of the flask
and the 400-ml mark The 100-ml and 300-ml must be closer to the 200-ml mark
than the bottom and 400-ml mark, respectively.
400 ml
300 ml
200 ml
100 ml
Florence Flask
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
45. . This question assesses the student’s ability to apply volume formulas and the concept of density in
modeling situations.
mass
volume
0.06 mg

V
0.06 mg

mg
1.2
mm3
 0.05 mm3
density =
1.2
(a)
mg
mm3
V
V
Photoresist
Silicon wafer
80 mm
(b) The photoresist covers the circular wafer with a uniform thickness, essentially making it a cylinder
with a height equal to the thickness:
V   r 2h
0.05 mm3   (40 mm) 2 h
0.05 mm3
 (40 mm) 2
h  0.000009947
h
h  10
5
mm
mm  10 nm
46. This questions assesses the student’s ability to
apply the area of triangles and quadrilaterals.
(a) Esmeralda county can be divided into a
triangle (south of dotted line) and a quadrilateral.
The quadrilateral north of the dotted line has an
area equal to the large triangle north of the dotted
line minus the small shaded triangle added to the
figure.
South triangle: base is about 70 miles and height
is about 60 miles. Area equals
1
 60 mi  70 mi   2100 mi2 .
2
North quadrilateral is the difference between the
large triangle with base 70 miles and height about
36 miles, and the shaded triangle with base and
height of 6 miles. Area equals
1
1
36 mi  70 mi    6 mi  6 mi   1242 mi2 .
2
2
The area of Esmeralda County is approximately
3342 square miles.
10 miles
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
(b) The population density of Esmeralda County is
775 people
 0.23 people per square mile.
3342 mi 2
47. This question assesses the student’s ability prove all circles are similar.
To prove circles A and B are similar, there must be a sequence of similarity
transformations that map circle A to circle B.
Translate circle A by vector AB . Under the translation, A = B.
Dilate circle A by a factor
b
. Since circle A was the set of all points distance a from
a
b
a
point A, circle A is the set of all points distance a   b from point A. That is the same set of points as
circle B. The composition of the transformation and dilation, both similarity transformations, maps circle
A to circle B. Thus, circles A and B are similar.
55. This question assesses the student’s ability to
construct a circle inside a triangle.
B
The student must (1) construct at least two angle
bisectors, (2) locate the point of intersection of the
bisectors (point O), (3) construct a perpendicular to
one side through O, (4) locate the point of
intersection of the side and the perpendicular line
(point P), and (5) construct a circle centered at O
whose radius is OP.
P
O
C
A
58. This question assesses the student’s ability to
construct a circle circumscribed about a triangle.
The student must (1) construct at least two perpendicular bisectors, (2) locate the point of
intersection of the bisectors (point O), and (3) construct a circle centered at O whose radius is OA
(or OB or OC).
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
72. This question assesses the student’s ability apply relationships between lines and angles in
circles.
Triangle AHO is a right triangle.
AO  2102  20002
AO  2011 km
The mountain’s height, AM = AO – 2000 km ≈ 11 km.
73. This question assesses the student’s ability to graph a circle, apply properties of parts of
circles, and use coordinate geometry to find equations of lines.
The radius shown has a slope of 1. Thus, the slope of the tangent line is –1.
The equation of the tangent line is y  4  1 x  4 or y   x  8 .
94. This question assesses the student’s ability to construct a tangent to a circle from a point.
The student must (1) construct AO , (2) construct the midpoint of AO at M, (3) construct a circle
of radius OM centered at M, (4) locate one intersection of circle M and circle O (point P), and
construct AP which is tangent to circle O.
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
95. This questions assesses the student’s understanding of the
relationship between the distance formula, Pythagorean Theorem,
and the equation of a circle.
y
Triangle ABC is a right triangle, so by the Pythagorean Theorem
AB2  AC 2  BC 2 .
B (x2, y2) or (x, y)
d
The distance d from A to B can be rewritten at
2
2
d 2   x2  x1    y2  y1  .
y2 – y1
A (x1, y1)
or (h, k) x2 – x1 C
The circle centered at A is the set of all points a distance d from
point A. Using B as an arbitrary point, the equation
2
2
d 2   x2  x1    y2  y1  holds. If points A and B are described as
having coordinates (h, k) and (x, y), respectively, and the circle is
described as having a radius of r, then r 2   x  h 2   y  k 2 .
x
102. This question assesses the student’s to graph linear functions, construct parallel and/or
perpendicular lines to given lines on the coordinate plane, and
y
compute the perimeter of a figure using coordinate geometry.

(a) See graph.
(b) A is (–6, 0) and B(6, 0). See graph.
m AP 
P
3.6  0
3.6 1


4.8   6  10.8 3
A
B

3.6  0 3.6

 3
4.8  6 1.2
 mAP  mBP    13   3  1
(c) mBP 
 x
Since the product of the slopes of AP and BP equal –1, the
segments are perpendicular and, therefore, APB is a right
angle. Thus, ABP is a right triangle.

104. This question assesses the student’s ability to apply circles and their equations in the coordinate plane.
(a) The most obvious choice is to center the circle at the midpoint of (1, 3) and (–5, 9), or (–2, 6). The radius of the
circle will be
2
1   2  3  62 
18 . The equation is then  x  2    y  6   18 . Any circle that goes
2
2
through the points (1, 3) and (–5, 9) is correct.
(b) Answers will vary depending on the equation, but obvious choices are y-intercepts
 0, 6 
 

14 or 0, 6  14 , or the lattice points 90° from the given points i.e. (1, 9) or (–5, 3).
106. This question assesses the student’s ability to apply the geometric definition of a parabola
on the coordinate plane.
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
The focus is p units from the vertex, therefore, points A and B have a y-coordinate equal to p.
p
1 2
x
4p
Thus, 4 p 2  x 2 . The distance from A(–2p, 0) to B(2p, 0) is 4p.
2 p  x
107. This question assesses the student’s ability to apply
the geometric definition of a parabola on the coordinate
plane.
y

(a) See graph.
P
(b) See graph. The focus is at (0, 1)
F
(c) See graph. Point P is at (3, 2.25).
PF 
(d)

 3  0 2   2.25  12
 3.25
PD  2.25   1
 x
D
. The distances are the same
 3.25
(which is how the parabola is defined.)

109. This questions assesses the student’s understanding of probability rules.
(a) P  A and B   P( A)  P(B)   0.3 0.4  0.12
(b) P  A or B   P( A)  (B)  P  A and B   0.3  0.4  0.12  0.58
(c) P  not A   1  P( A)  1  0.3  0.7
(d) Since A and B are independent, P  A | B   P  A  0.3
124. This questions assesses the student’s ability to compile data into a two way table and apply
rules of probability, and understanding of the concept of independence.
(a) See table.
GEOMETRY HONORS
2014-2015 SEMESTER EXAMS
PRACTICE MATERIALS KEY
SEMESTER 2
Selected Response Key
(b, i) P(siblings yes) 
21 3

35 5
(b, ii) P(siblings yes or pets yes) 
(b, iii) P(siblings yes | pets yes) 
21 21 15 27
  
35 35 35 35
15 5

21 7
(e) No, because
P(siblings yes)  P(siblings yes | pets yes)
3 5

5 7
128. . This questions assesses the student’s ability to
apply rules for probability to compound events and
real-world situations.
This question can be solved using multiple methods,
including formulas, a tree diagram, or a table
(as shown). Values based on given information is bold.
(a) P(Classified as spam?)  16.25%
(b) P(Spam | Classified spam) 
12%
 74%
16.25%
(c) P  Legitimate and classified Spam  P(Spam and classified Legitimate)  4.25%  3%  7.25%
(d) P(Classified Spam | Legitimate)  5% (given)
P( Not Classified Spam | Spam) 
3%
 20% . It is 4 times more likely that a spam message will get
15%
through the filter than a legitimate message will be classified as spam.